Two theories:
- You take away one option, you are left with two options. It must be 50-50, right?
- You have a 1/3 chance it's behind the first door and that doesn't change just because it's not behind the third door.
I will go to my grave thinking that the first theory makes the most sense. What if you started out with two doors, had the contestant chose one of them, then wheeled out a third door, opened it and there was a goat there. Would that have made it a good idea for the contestant to switch? Isn't that the same situation?
I decided to test it with a deck of cards. I separated the black and red cards. The black cards are cars, the red goats. I combined one black card with two red cards. I shuffled the three cards as best I could. I took the first three and put them face down. I 'chose' #1 and flipped the third card over. if it was black, I put those three cards aside. If it was red, I then flipped the second card and recorded when it was black and when it was red. I had, as I recall 10 of 13 sets where the third card was red and the second card was black 7 times. I reshuffled them and did it a gain a couple more times and the 2/3 percentage held. I then realized that I was wrong but still don't understand why.
But if Lampkin stays, I'll still be happy.